Hypothesis tests

Appendices

Chi-Square Goodness of Fit Test

This lesson explains how to conduct a
chi-square goodness of fit test.
The test is applied when you have one
categorical variable
from a single population. It is used to determine whether sample
data are consistent with a hypothesized distribution.

For example, suppose a company printed baseball cards. It
claimed that 30% of its cards were rookies; 60% were veterans but not All-Stars; and
10% were veteran All-Stars. We could gather a random sample of baseball
cards and use a chi-square goodness of fit test to see whether
our sample distribution differed significantly from the
distribution claimed by the company. The
sample problem at the end of the lesson
considers this example.

When to Use the Chi-Square Goodness of Fit Test

The chi-square goodness of fit test is appropriate when the
following conditions are met:

The expected value of the number of sample observations in each
level
of the variable is at least 5.

This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst
to state a
null hypothesis (H0)
and an
alternative hypothesis (Ha). The hypotheses are stated in such
a way that they are mutually exclusive. That is, if one is
true, the other must be false; and vice versa.

For a chi-square goodness of fit test, the hypotheses take
the following form.

H0: The data are consistent with a specified distribution.

Ha: The data are not consistent with a
specified distribution.

Typically, the null hypothesis (H0) specifies the proportion of
observations at each level of the categorical variable. The
alternative hypothesis (Ha) is that at least one of
the specified proportions is not true.

Formulate an Analysis Plan

The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. The plan should specify the following elements.

Significance level. Often, researchers choose
significance levels
equal to
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.

Test method. Use the
chi-square goodness of fit test
to determine whether
observed sample frequencies differ
significantly from expected frequencies specified in the
null hypothesis. The chi-square goodness of fit test is
described in the next section, and demonstrated in the sample
problem at the end of this lesson.

Analyze Sample Data

Using sample data, find the
degrees of freedom, expected frequency counts,
test statistic, and the P-value associated with the test statistic.

Degrees of freedom. The
degrees of freedom (DF) is equal to the
number of levels (k) of the categorical variable minus 1.

DF = k - 1

Expected frequency counts. The expected frequency counts
at each level of the categorical variable are equal to
the sample size times the hypothesized proportion
from the null hypothesis

Ei = npi

where
Ei is the expected frequency count for the
ith level of the categorical variable,
n is the total sample size, and
pi is the hypothesized proportion of observations
in level i.

Test statistic. The test statistic is a chi-square random variable
(Χ2) defined by
the following equation.

Χ2 =
Σ [ (Oi - Ei)2 / Ei ]

where
Oi is the observed frequency count for the
ith level of the categorical variable, and
Ei is the expected frequency count for the
ith level of the categorical variable.

P-value. The P-value is the probability of observing a
sample statistic as extreme as the test statistic. Since the
test statistic is a chi-square, use the
Chi-Square Distribution Calculator
to assess the probability associated with the test statistic. Use
the degrees of freedom computed above.

Interpret Results

If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the P-value to the
significance level,
and rejecting the null hypothesis when the P-value is less than
the significance level.

Test Your Understanding

Problem

Acme Toy Company prints baseball cards. The company claims that 30% of
the cards are rookies, 60% veterans but not All-Stars, and 10% are veteran All-Stars.

Suppose a random sample of 100 cards has 50 rookies, 45
veterans, and 5 All-Stars. Is this consistent with Acme's claim?
Use a 0.05 level of significance.

Solution

The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:

State the hypotheses. The first step is to
state the null hypothesis and an alternative hypothesis.

Null hypothesis: The proportion of rookies, veterans, and
All-Stars is 30%, 60% and 10%, respectively.

Alternative hypothesis: At least one of the proportions in the
null hypothesis is false.

Formulate an analysis plan. For this analysis,
the significance level is 0.05. Using sample data, we will
conduct a
chi-square goodness of fit test
of the null hypothesis.

Analyze sample data. Applying the chi-square
goodness of fit test to sample data, we compute
the degrees of freedom,
the expected frequency counts, and
the chi-square test statistic. Based on the
chi-square statistic and the
degrees of freedom, we determine the
P-value.

where
DF is the degrees of freedom,
k is the number of levels of the categorical variable,
n is the number of observations in the sample,
Ei is the expected frequency count for level i,
Oi is the observed frequency count for level i, and
Χ2 is the chi-square test statistic.

The P-value is the probability that a chi-square statistic
having 2 degrees of freedom is more extreme than 19.58.

Interpret results. Since the P-value (0.0001) is
less than the significance level (0.05), we cannot accept the
null hypothesis.

Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling, the
variable under study was categorical, and each level of the categorical
variable had an expected frequency count of at least 5.